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Likurg_2 [28]
3 years ago
13

The least common multiple of two numbers is 60, and one of the numbers is 7 less than the other number. What are the numbers? Ju

stify your answer
Mathematics
1 answer:
Aleks [24]3 years ago
6 0
To do this we need to find the factors of 60 these are:
1 and 60
2 and 30
3 and 20
4 and 15
5 and 12
6 and 10
There is 1 pair that have a difference of 7 and that is 5 and 12
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Pick your answer is it <br> A <br> B <br> C or <br> D
aleksley [76]

Answer:

I belive its B,C

sorry if i'm wrong but hope it helps

Step-by-step explanation:

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3 years ago
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Geometry Pythagorean theorem
lara31 [8.8K]

Answer:

Below!

Step-by-step explanation:

Using Pythagoras theorem, I will solve all of the problems.

<h3>________________________________________________</h3>

<u>Question 9:</u>

  • 10² = 6² + x²
  • => 100 = 36 + x²
  • => 100 - 36 = x²
  • => 64 = x²
  • => x = 8
<h3>________________________________________________</h3>

<u>Question 10:</u>

  • 26² = 24² + x²
  • => 676 = 576 + x²
  • => 676 - 576 = x²
  • => 100 = x²
  • => x = 10
<h3>________________________________________________</h3>

<u>Question 11:</u>

  • 15² = 12² + x²
  • => 225 = 144 + x²
  • => 225 - 144 = x²
  • => 81 = x²
  • => x = 9
<h3>________________________________________________</h3>

<u>Question 12:</u>

  • x² = 8² + 12²
  • => x² = 64 + 144
  • => x² = 208
  • => x = √208
  • => x = 14.2 (Rounded)
<h3>________________________________________________</h3>

<u>Question 13:</u>

  • 7² = 2² + x²
  • => 49 = 4 + x²
  • => 49 - 4 = x²
  • => 45 = x²
  • => x = √45
  • => x = 6.7 (Rounded)
<h3>________________________________________________</h3>

<u>Question 14</u>

First, let's solve for the variable x using Pythagoras theorem.

  • => 5² = 3² + x²
  • => 25 = 9 + x²
  • => 16 = x²
  • => x = 4 units

Now, let's solve for the variable y using Pythagoras theorem.

  • (3 + 6)² = 5² + y²
  • => (9)² = 25 + y²
  • => 81 = 25 + y²
  • => y² = 56
  • => y = √56
  • => y = 7.5 (Rounded) units

Answers (Nearest tenth):

  • x = 4 units
  • y = 7.5 units
<h3>________________________________________________</h3>

<u>Question 15:</u>

First, let's find the value of the variable y using Pythagoras theorem.

  • 8² = 6² + y²
  • => 64 = 36 + y²
  • => 28 = y²
  • => y = √28
  • => y = 5.3 (Rounded) units

Now, let's find the value of the variable x using multiplication.

  • x = 2y
  • => x = 2(5.3)
  • => x = 10.6 units

Answer (Nearest tenth)

  • x = 10.6 units
  • y = 5.3 units
<h3>________________________________________________</h3>
5 0
2 years ago
Solve the absolute value inequality.
Alex Ar [27]

Answer: k ≤ -4 or k ≥ \frac{4}{3}

<u>Explanation:</u>

                      | 3k + 4 | ≥ 8

3k + 4 ≥ 8             or                3k + 4 ≤ -8

<u>       -4</u>  <u> -4  </u>                              <u>       -4</u>    <u>-4 </u>

3k       ≥ 4              or                3k       ≤ -12

<u>÷3       </u> <u>÷3  </u>                              <u> ÷3       </u>   <u>÷3 </u>

k        ≥ \frac{4}{3}            or                 k       ≤  -4

4 0
3 years ago
Wirk out the values of 5x-2y when x=-2 and y = -2
elena-s [515]

Step-by-step explanation:

5x - 2y

5(-2) - 2(-2)

-10 + 4

= - 6

3 0
3 years ago
Can I get some help, it is due in 15 minutes and I need help. Thanks, any help appreciated
koban [17]

Well, I'm way past the 15 min mark, but here's how to do the question.


With this, you will need to use the distance formula, \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}, on XY, YZ, and ZX.



XY: \sqrt{(3-1)^2+(1-6)^2}


Firstly, solve inside the parentheses: \sqrt{(2)^2+(-5)^2}


Next, solve the exponents: \sqrt{4+25}


Next, solve the addition, and XY's distance will be √29



(The process is the same with the other 2 sides, so I'll go through them real quickly)


YZ:

\sqrt{(6-3)^2+(3-1)^2}\\ \sqrt{(3)^2+(2)^2}\\ \sqrt{9+4}\\ \sqrt{13}



ZX:

\sqrt{(1-6)^2+(6-3)^2}\\ \sqrt{(-5)^2+(3)^2}\\ \sqrt{25+9}\\ \sqrt{34}



Now that we got the 3 sides, we can add them up: \sqrt{29}+\sqrt{13} +\sqrt{34} =14.8


In short, your answer is 14.8, or the second option.

8 0
3 years ago
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